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7 days ago

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A kettle heats 1.75 kg of water. The specific latent heat of vaporisation of water is 3.34 x 106 J/kg. How much energy would be

needed to boil off the water? Write the answer out in full (i.e. not in standard form).

1 answer:

Softa [913]7 days ago

7 0

The total energy required to evaporate the water is 5,845,000 J

Explanation:

The quantity of thermal energy needed to entirely vaporize a specific amount of a liquid substance that is already at its boiling point is represented by

$Q=m\lambda_v$

where

m indicates the mass of the substance

$\lambda_v$ represents the specific latent heat of vaporization for that substance

In this scenario, the substance in question is water, with a mass of

m = 1.75 kg

The specific latent heat of vaporization for water is

$\lambda_v = 3.34\cdot 10^6 J/kg$

Assuming the water has reached its boiling point, the thermal energy necessary to vaporize it is

$Q=(1.75)(3.34\cdot 10^6)=5,845,000 J$

Discover more about specific heat:

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A locomotive is accelerating at 1.6 m/s2. it passes through a 20.0-m-wide crossing in a time of 2.4 s. after the locomotive leav

kicyunya [1025]

Response:

Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.

Details:

The locomotive's acceleration is 1.6 $m/s^2$

The duration taken to pass the crossing is 2.4 seconds.

We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.

When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 $m/s^2$ .

32 = 0 + 1.6 * t

t = 20 seconds.

Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.

Thus, after clearing the crossing, it takes an additional 17.6 seconds to reach the speed of 32 m/s.

6 0

5 days ago

Read 2 more answers

A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro

Ostrovityanka [942]

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

According to the problem, the distance from the building where the ball hits is 16m, and its final elevation exceeds the initial height by 8m.

With this information, we can compute the ball’s starting speed.

a) Let's first assess the horizontal trajectory.

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

This gives us our initial equation.

Next, we need to examine the vertical trajectory.

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Utilizing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Now let’s solve for t.

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

The ball takes two seconds to reach the adjacent building, allowing us to compute its initial speed.

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To determine the velocity magnitude just before impact, we must calculate both x and y components.

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

The computed velocity magnitude is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The ball's angle is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

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13 days ago

Ben walks 500 meters from his house to the corner store. He then walks back toward his house, but continues 200 meters past his

Keith_Richards [1034]

Velocity = 71 meters per minute (MPM)
S stands for Speed
D means Distance
T represents Time
To calculate Speed, divide Distance by Time.

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14 days ago

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kicyunya [1025]

Answer:

Maximum emf = 5.32 V

Explanation:

Provided data includes:

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions each second

Magnetic field, B = 0.5 T

We are tasked to determine the maximum emf produced in the loop, which is founded on Faraday's law. The induced emf can be calculated by:

$\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t$

For the maximum emf, $\sin\omega t=1$

Therefore,

$\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V$

Hence, the maximum emf generated in the loop is 5.32 V.

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A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth

serg [1198]

The kangaroo reaches a maximum vertical altitude of 2.8 m, which can be calculated using the formula 2.8 = 1/2 * 9.8 * t^2. Thus, applying the equation s = ut + 1/2at^2.

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